The equation of the circle passing through `(2,0)` and `(0,4)` and having minimum radius is

Options

A. `x^(2)+y^(2)=20`
B. `x^(2)+y^(2)-2x-4y=0`
C. `(x^(2)+y^(2)-4)+lamda(x^(2)+y^(2)-16)=0`
D. None of the above

Answer

To find the equation of the circle that passes through the points (2, 0) and (0, 4) and has the minimum radius, we can follow these steps:

### Step 1: Write the general equation of the circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((-g, -f)\) is the center of the circle and \(r\) is the radius.

### Step 2: Substitute the first point (2, 0)
Substituting the point (2, 0) into the general equation:
\[ (2)^2 + (0)^2 + 2g(2) + 2f(0) + c = 0 \]
This simplifies to:
\[ 4 + 4g + c = 0 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (0, 4)
Now, substitute the point (0, 4) into the general equation:
\[ (0)^2 + (4)^2 + 2g(0) + 2f(4) + c = 0 \]
This simplifies to:
\[ 16 + 8f + c = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve for \(c\) from both equations
From Equation 1:
\[ c = -4 - 4g \]
From Equation 2:
\[ c = -16 - 8f \]
Setting the two expressions for \(c\) equal:
\[ -4 - 4g = -16 - 8f \]
Rearranging gives:
\[ 8f - 4g = -12 \quad \text{(Equation 3)} \]

### Step 5: Express \(g\) in terms of \(f\)
From Equation 3, we can express \(g\):
\[ 4g = 8f + 12 \implies g = 2f + 3 \]

### Step 6: Find the radius in terms of \(f\)
The radius \(r\) of the circle can be expressed as:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting for \(g\) and \(c\):
\[ r = \sqrt{(2f + 3)^2 + f^2 - (-16 - 8f)} \]
This simplifies to:
\[ r = \sqrt{(4f^2 + 12f + 9) + f^2 + 16 + 8f} \]
\[ r = \sqrt{5f^2 + 20f + 25} \]
\[ r = \sqrt{5(f^2 + 4f + 5)} = \sqrt{5} \sqrt{(f + 2)^2 + 1} \]

### Step 7: Minimize the radius
The minimum radius occurs when the term \(\sqrt{(f + 2)^2 + 1}\) is minimized. This term is minimized when \(f + 2 = 0\) or \(f = -2\).

### Step 8: Substitute \(f\) back to find \(g\) and \(c\)
Substituting \(f = -2\) into \(g = 2f + 3\):
\[ g = 2(-2) + 3 = -4 + 3 = -1 \]
Now substituting \(f = -2\) into \(c = -16 - 8f\):
\[ c = -16 - 8(-2) = -16 + 16 = 0 \]

### Step 9: Write the final equation of the circle
Now substituting \(g\), \(f\), and \(c\) back into the general equation:
\[ x^2 + y^2 + 2(-1)x + 2(-2)y + 0 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x - 4y = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 2x - 4y = 0 \]